(a) For any real show that What does this tell you about the given
(b) Show that . Compare and
(c) Combine conclusions of parts a and b to express and therefore the desired quantity in a suitable form.
Solution
(a) Because of the absolute value we may assume that by replacing with if necessary. Now use AM-GM inequality or the fact that since given must be a non-real complex number.
(b) Let Then Using the formulas for and canceling gives the quadratic equation since , we get
(c) Let Then Adding and using that we get and So .
Set of powers
A5, 2021
For a complex number let . (a) For every natural number there exits a such that . (b) There exits a unique such that . (c) If and then is an infinite set. (d) has infinite elements.
Solution (a) True. Pick . (b) False. Both and work. (c) True. If , then every number in has a different length. (d) True. If for some powers with , then for some natural number . But this would imply that is rational.
Complex polygon
B2, 2020
2. (i) Let where is a positive integer. Prove that
Solution
Since , we have .
For , . So the first factor must be zero. This proves the statement.
2. (ii) Prove that
Solution
Notice that . Let . Then:
From problem 2(i), we know that RHS of the above equation is zero. Since and are real numbers, both of them must be individually zero. In particular, , which proves the statement.
Absolute values
A9, 2022
Let be a non-real complex number and . Which of the statements are correct?
If , then must be real.
If , then .
If is real, then .
If is real, then .
Solution
TRUE.
TRUE.
TRUE.
FALSE.
Power of a complex number
A13, 2010
If is a real number satisfying find the value of where
Solution
Let us know look at the quantity we want to compute:
Complex triangle
A7, 2013
Let be angles such that form an equilateral triangle in the complex plane. Find the values of the given expressions.
a)
b)
c)
d)
Solution
a) by taking the vector sum of the three points on the unit circle.
b) real part of which is 0 by part a.
c) because the points on the unit circle also form an equilateral triangle in the complex plane, since taking , we get and . The last term does not change the position of the point.
d) because, using the formula for in part we get and the sum of the LHS and the RHS in this equation is
Maximum and minimum of an average
A9, 2014
Let be real numbers and let be the average of the complex numbers where . As the values of 's vary over all 13 -tuples of real numbers, find
(i) the maximum value attained by
(ii) the minimum value attained by .
Solution
(i) Each can take a maximum value of 1, which is attained when . Hence, the maximum average is also 1.
(ii) To get the minimum, place the 13 points on the vertices of a regular cyclic polygon. The average corresponds to the center of the polygon which is . Hence, the minimum value can take is 0.
Roots of unity I
A10, 2015
(i) Suppose are complex numbers. One of them is in the second quadrant and the other is in the third quadrant. How does compare with ?
(ii) Complex numbers and 0 form an equilateral triangle. How does compare with.
(iii) Let be the complex 8 -th roots of unity. Find the value of where the symbol denotes product.
Solution
(i) || One way: using triangle inequality for and we get and so Now we may take absolute value on the LHS because switching and keeps RHS the same. For equality, and must point in the same direction, so and must be along the same line. But they are in quadrants 2 and so this cannot happen.
(ii) must be obtained by rotating by angle say in the counterclockwise direction (otherwise interchange the two). Then . Then and Now (see by calculation or picture), so we have in fact
(iii) We have . Putting gives
Counting roots in a quadrant
A6, 2018
Consider the equation
where
(a) How many complex solutions does this equation have?
(b) How many solutions does each of the four quadrants have?
Solution
(a) In general, the equation has solutions given by: The given equation has 2018 complex solutions, since we can express the complex number in the RHS as for some small .
(b) Instead of looking at the given equation, first consider the solutions to: . Two of them are real values: and . The other 2016 are distributed equally in the four quadrants, 504 each.
If we rotate the solutions to by a tiny angle in the counter-clockwise direction, we get the solutions to the given equation. (The value of being ). This gives 505 values each in the first and third quadrant but still 504 in the second and fourth quadrant.